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Cp - cpk for screwdriving process

Discussion in 'SPC - Statistical Process Control' started by MarkusF, May 2, 2024.

  1. MarkusF

    MarkusF New Member

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    Hi to all,
    I am struggling with the validation of a machine that tightens 5 screws (same screws with the same tightening parameters) on our product. Is it sufficient to validate the cpk and cp only on one of the 5 screws?
    Thank you very much.
     
  2. Miner

    Miner Moderator Staff Member

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    But does the machine use the same driver for all 5 screws? Or does it use 5 different drivers? Does it collect the torque data (like an Atlas-Copco system)? Then it would be easy to determine capability for all 5.
     
  3. MarkusF

    MarkusF New Member

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    Same driver for all 5 screws. With same parameter. So I think that capability for only 1 screw make sense, right ? It does not collect any datalog maybe in the future.
     
  4. Miner

    Miner Moderator Staff Member

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    If you are going to perform a 100-piece capability study, I recommend measuring 20-pieces from each screw. Plot the data on an I-MR control chart and look for any signs of an out-of-control process, particularly shifts in the mean when transitioning from one screw to the next. I would also run an ANOM (Analysis of Means) on the data to see if that shows a difference between screws. If the process is in control and the ANOM does not show a difference, you can combine the results in your capability study AND be confident that there are no differences.
     
    Bev D likes this.
  5. MarkusF

    MarkusF New Member

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    Unfortunately I have only about 30 parts, i am not an expert of statistics i'm an electronic.
    I'm sorry for that. I'm not understanding why i can't make an easy validation using the values of only 1 screw if there is only 1 screwdriver and the screws will be screwed with the same parameters.

    What is the difference beetween this two kind of approach and why should be take in exam all the screws ?

    Thanks a lot.
     
    Last edited: May 3, 2024
  6. Miner

    Miner Moderator Staff Member

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    It may not make a difference, but then again it may. A simple study such as I described would answer that question definitively. I've seen a lot a strange things in my career that flew in opposition to conventional wisdom.
     
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  7. Bev D

    Bev D Moderator Staff Member

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    What Miner said.

    The problem here as that the screws which are in different positions may be different because of the location - this is called homogeneity. If they are homogenous you can combine the results. (This is what Miner is asking you to test for.). If they not homogenous (different average value from each other) you cannot directly combine them because the difference is ‘fixed’. If you were to combine them then the fixed difference would be seen mathematically as if it were random but it’s not - the formula will give you a invalid result - and more importantly you would be missing a significant source of variation.

    Statistics is just as complex and deep as electronic engineering - I know, I have degrees and 40 years of experience in both. I too have seen crazy things: once I had 4 machines that were substantially different from each other although they were the same brand, model and their SNs were sequential. The difference? The location changed the hydraulic pressures just enough because the lines were different lengths.

    So, unless you want an in depth statistics lesson, you should follow Miner’s advice. It would be helpful to post your results here so we can help you interpret the results. Statistics isn’t a shake and bake oven…;)
     
    Last edited by a moderator: May 3, 2024
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  8. MarkusF

    MarkusF New Member

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    Thanks to all for the support. I'm going to decide what do to in the next months but I am happy because i found a very useful forum ! Thanks a lot.
     
    Andy Nichols likes this.